Answer to Question #104033 in Classical Mechanics for Mkt

Question

P=12N, Q=5N, theta=120 degrees
Find the resultant force and alpha

Answer 1

As per the given question,

P=12N

Q=5N

Angle between the force $\theta=120^\circ$

$R=\sqrt{P^2+Q^2+2PQ\cos\theta}$

$R=\sqrt{12^2+5^2+2\times12\times5\cos(120^\circ)}$

$R=\sqrt{144+25+120\cos{(90^\circ+30^\circ)}}$

$R=\sqrt{169-120\sin30^\circ}$

$R=\sqrt{169-120\dfrac{1}{2}}$ N

$R=\sqrt{169-60}=\sqrt{109}$